Some individuals produce more offspring than others.
If you have the alleles A and a, then the The hardy-weinberg equation is: A2 + 2Aa + a2 = 1 where a and A represent allele frequencies. So A2 would be the genotype frequency for AA. 2Aa is the genotype frequency for Aa. And a2 is the genotype frequency for aa. Plug in whatever information you have into the equation and you can probably come up with an answer. Save
To calculate q, you first need to determine the frequency of the recessive allele in the population, which can be done using the Hardy-Weinberg equation. In this case, since 11 out of 100 individuals are showing the recessive trait (green color), the q² value is 0.11 (11/100). Therefore, q, the frequency of the recessive allele, would be the square root of 0.11, which is approximately 0.33.
If heterozygous individuals are not favored, then the frequency of heterozygous individuals will decrease as the frequency of homozygous individuals increase. This can be shown using the Hardy-Weinberg equation for allele frequencies in a population: p2 + 2pq + q2 = 1 where q2 & p2 are the frequencies of the two different homozygous individuals (eg. aa and AA) and 2pq is heterzygous (eg. Aa). As the equation shows, if 2pq decreases, the other two variables must increase to compensate.
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Yes, a gas trap can be an accurate measure of decarboxylation of an amino acid since the evolution of carbon dioxide is a characteristic reaction of decarboxylation. By capturing and measuring the evolved carbon dioxide in the gas trap, one can infer the extent of decarboxylation that has occurred.
The Hardy-Weinberg law assumes that there is no mutation occurring in the population because mutations can introduce new alleles, disrupting the equilibrium between allelic frequencies. Including mutations would complicate the predictive power of the Hardy-Weinberg equilibrium.
The five Hardy-Weinberg principles are:1. No mutations2. No natural selection3. random mating4. a large population5. no immigration or emigrationIt is impossible to have no natural selection in a natural environment because that would require all organisms to be equally fit. The only way to meet this principle is to have a population of genetically identical organisms which does not happen naturally. All five of these principles cannot be met in real life, but it may be possible to have a species in hardy- weinberg equilibrium in a lab situation.
This would be having exactly enough, but not too much of the product in demand. So you would be maximizing profit!
The Hardy-Weinberg equilibrium can be disrupted by deviations from any of its five main underlying conditions. Therefore mutation, gene flow, small population, nonrandom mating, and natural selection will disrupt the equilibrium.
Very large population, isolation from other population, no net mutation, random mating, no natural selection...if these conditions are met, the status would be in equilibrium or in other words there is no change.
genotype frequencies in a population over multiple generations. It provides a baseline expectation under certain assumptions, such as no selection, mutation, migration, or genetic drift, allowing researchers to infer mechanisms shaping genetic variation. Deviations from these expectations can signify evolutionary forces at play within a population.
the reaction would shift to favor the side with the most moles of gas
No Jeff hardy was never gay. If he was gay or is you would be able to tell.
Probably, but it would cost like $300,000.
Possibly. If there are no external forces acting upon it, and hence no acceleration, then yes, the body can be in equilibrium. But if there is no acceleration,would the body be in motion?
If he liked you!
jeff hardy wind 3 or 2 times